Question

\(\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx=sin3x\)

Answer 1

`\sqrt{3}/2 cos x-1/2 sin x=sin 3x`

`<=>sin` `\pi/3 cos x-cos` `\pi/3 sin x=3x`

`<=>sin(\pi/3-x)=3x`

`<=>` $\left[\begin{matrix} \dfrac{\pi}{3}-x=3x+k2\pi\\ \dfrac{\pi}{3}-x=\pi-3x+k2\pi\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\ x=\dfrac{\pi}{3}+k\pi\end{matrix}\right.$   `(k in ZZ)`

Vậy ptr có `2` họ nghiệm `x=\pi/12-[k\pi]/2` ; `x=\pi/3+k\pi`  `(k in ZZ)`